Question

How do you evaluate `[cos(arccos (3/5) - arcsin (4/5))]`?

Answer 1

`arccos(3/5) = arcsin(4/5)` so this is just `cos(0) = 1`

Explanation:

`3^2+4^2=5^2`

So a `3,4,5` triangle is a right angled triangle.

`arccos(3/5) = B = arcsin(4/5)`

So: `cos(arccos(3/5)-arcsin(4/5)) = cos(0) = 1`